Problem 79 - Calculus for Biology and Medicine

Question

Simplify the following expressions: (a) \(-\ln \frac{1}{3}\) (b) \(\log _{4}\left(x^{2}-4\right)\) (c) \(\log _{2} 4^{3 x-1}\)

Step-by-Step Solution

Answer

(a) \(\ln 3\); (b) \(\log_4(x-2) + \log_4(x+2)\); (c) \(6x - 2\).

Step 1: Understanding Expression (a)

To simplify \(-\ln \frac{1}{3}\), let's begin by using the logarithm property \(\ln \frac{1}{a} = -\ln a\). This gives us: \\[-(-\ln \frac{1}{3}) = \ln 3\]

Step 2: Understanding Expression (b)

To simplify \(\log _{4}\left(x^{2}-4\right)\), notice that \(x^{2} - 4 = (x - 2)(x + 2)\), which is a difference of squares factorization. Thus, we have: \\[\log _{4}(x^{2} - 4) = \log _{4}((x - 2)(x + 2))\]\Using the property of logarithms that states \(\log_b(mn) = \log_b m + \log_b n\), we get: \\[\log _{4}(x - 2) + \log _{4}(x + 2)\]

Step 3: Understanding Expression (c)

To simplify \(\log _{2} 4^{3 x-1}\), use the exponentiation property of logarithms, \(\log_b(a^n) = n\cdot\log_b a\), with \\[n = 3x - 1, \, a = 4\]\.Since \(4 = 2^2\), we can write: \\[[3x - 1]\cdot\log_2(2^2) = [3x - 1]\cdot 2\cdot \log_2 2\]\Noting that \(\log_2 2 = 1\), this simplifies to \\[2(3x - 1) = 6x - 2\]

Key Concepts

Logarithmic Properties

Logarithmic properties are fundamental rules that govern how logarithms behave. These properties help manipulate and simplify logarithmic expressions. Let's go through some essential properties:

• The Product Property: \(\log_b(mn) = \log_b m + \log_b n\). This property states that the logarithm of a product is the sum of the logarithms.
• The Quotient Property: \(\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n\). The logarithm of a quotient is the difference of the logarithms.
• The Power Property: \(\log_b(a^n) = n \cdot \log_b a\). It shows that the logarithm of a power can be brought outside as a coefficient.
• The Change of Base Formula: \(\log_b a = \frac{\log_k a}{\log_k b}\). This formula is helpful when converting logs from one base to another.

Applying these rules helps break down complex logarithmic expressions into simpler components. For instance, in expression (a) from the exercise, we used \(\ln \frac{1}{a} = -\ln a\) to simplify to \(\ln 3\). Understanding these properties makes working with logs much easier.

Simplification of Expressions

Simplification of expressions often involves recognizing patterns or identities that allow us to rewrite expressions more conveniently. This process is important as it makes expressions easier to evaluate or manipulate further.
One classic method of simplification is factoring. In expression (b), \(\log_4(x^2 - 4)\), the expression was simplified using the difference of squares formula. Recognizing that \(x^2 - 4\) can be factored into \((x - 2)(x + 2)\) allowed us to use the Product Property of logarithms: \(\log_b(mn) = \log_b m + \log_b n\). This took a compound expression and made it the sum of two simpler logarithms.
Another approach to simplification is using logarithmic identities to manipulate and consolidate terms. Simplifying log expressions requires careful observation and strategic application of different algebraic identities. This step-by-step approach ensures clarity and accuracy.

Algebraic Manipulation

Algebraic manipulation involves rewriting an expression without changing its value to achieve a simpler form. This involves several techniques including factoring, distribution, and substitution.
In expression (c), \(\log_2 4^{3x-1}\), the use of algebraic manipulation was evident. First, consider the expression with exponential terms. Recognize that 4 is equal to \(2^2\). Using the Power Property, \(\log_b(a^n) = n \cdot \log_b a\), helps us simplify the expression considerably.
The expression \((3x - 1)\log_2(4)\) was modified by expressing 4 as a power of 2: \(2^2\). By multiplying the exponent through, with \(\log_2 2 = 1\), the expression is simplified to a straightforward linear form \(6x - 2\). Through algebraic manipulation, sometimes expressions become much shorter and simpler, facilitating easier calculation and interpretation.
Honing algebraic manipulation skills is essential for efficiently solving complex equations in both algebra and calculus.